QJEEM22Ma05ff: JEE Main 2022 Mathematics - Class 12 JEEM22Ma05ff Solution: Let the solution curve of the differential equation , be . Then …

Given,

$x{{dy} \over {dx}} - y = \sqrt {{y^2} + 16x}$

$\Rightarrow x{{dy} \over {dx}} = y + \sqrt {{y^2} + 16x}$

$\Rightarrow {{dy} \over {dx}} = {y \over x} + \sqrt {{{\left( {{y \over x}} \right)}^2} + 16}$

This is a homogenous different equation.

Let ${y \over x} = v$

$\Rightarrow y = vx$

$\Rightarrow {{dy} \over {dx}} = v + x{{dv} \over {dx}}$

$\therefore$ $v + x{{dv} \over {dx}} =v+ \sqrt {{v^2} + 16}$

$\Rightarrow$ $x{{dv} \over {dx}} = \sqrt {{v^2} + 16}$

$\Rightarrow {{dv} \over {\sqrt {{v^2} + 16} }} = {{dx} \over x}$

Integrating both sides, we get

$\int {{{dv} \over {\sqrt {{v^2} + 16} }} = \int {{{dx} \over x}} }$

$\Rightarrow \ln \left| {v + \sqrt {{v^2} + 16} } \right| = \ln x + \ln c$

$\Rightarrow v + \sqrt {{v^2} + 16} = cx$

Now putting, $v = {y \over x}$ , we get

${y \over x} + \sqrt {{{{y^2}} \over {{x^2}}} + 16} = cx$

$\Rightarrow {y \over x} + \sqrt {{{{y^2} + 16{x^2}} \over {{x^2}}}} = cx$

$\Rightarrow y + \sqrt {{y^2} + 16{x^2}} = c{x^2}$ ...... (1)

Given, $y(1) = 3$

$\therefore$ When x = 1 then y = 3.

Putting in equation (1) we get,

$3 + \sqrt {9 + 16} = c.\,1$

$\Rightarrow c = 8$

$\therefore$ Solution of equation,

$y + \sqrt {{y^2} + 16{x^2}} = 8{x^2}$

Now, y(2) means when x = 2 then y = ?

$\therefore$ $y + \sqrt {{y^2} + 16 \times 4} = 8 \times 4$

$\Rightarrow y = 15$